Group Theory. Ferdi Aryasetiawan Dept. of Theoretical Physics, University of Lund, Sölvegatan 14A, Lund - Sweden

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1 Group Theory Ferdi Aryasetiawan Dept. of Theoretical Physics, University of Lund, Sölvegatan 14A, Lund - Sweden

2 Contents 1 Abstract Group Theory Group Abelian Group Subgroup Examples Multiplication Table Cyclic Groups Order of an Element Properties of Finite Groups Cosets Class Isomorphism and Homomorphism Invariant Subgroup Direct Product Group Theory of Group Representations Definitions Equivalent Representations Reducible and Irreducible Representation Equivalent Unitary Representation Schur s Lemma Orthogonality Theorem The Characters of a Representation The Regular Representation The Character Table Properties of the Character Table Class Multiplication i

3 ii CONTENTS 2.12 Direct Product Groups Examples Group Theory in Quantum Mechanics Linear Vector Space Symmetry Transformations Invariant Subspace and Its Generation Basis Functions for a Representation Projection Operators Orthogonality of Basis Functions Examples Relationship to Quantum Mechanics The Group of the Schrödinger Equation Degeneracy and Invariant Subspace Partial Diagonalisation of Ĥ Irreducible Sets of Operators Direct Product Representations of a Group Clebsch-Gordan Coefficients The Wigner-Eckart Theorem Applications Examples Lie Groups Definitions Infinitesimal Operators Lie Algebra Casimir Operators Ladder Operators and Multiplets Summation over group elements Rotation Group R Infinitesimal operators Commutation Relations Casimir Operator Irreducible Representations of R Characters of R The Vector-Coupling Theorem Clebsch-Gordan Coefficients

4 CONTENTS iii 4.12 Spherical Harmonics Atomic Physics Ĥ Ĥ 0 + Ĥ Configuration and Term Ladder Operator Method Projection Operator Method Term Energies Ĥ = Ĥ0 + Ĥ1 + Ĥ The Group SU 2 : Isospin The Group SU Relationship between SU 2 and R SU 2 in Nuclei Tensor Operators The Point Groups Crystal Symmetry Schoenflies Notation Commuting Operations Enumeration of Point Groups Improper Point Groups Double-Group Representations Space Group The Group SU Infinitesimal Operators and Lie Algebra Subgroups Multiplet Structure Example Product Representations SU 3 Multiplets and Hadrons Casimir Operators

5 Chapter 1 Abstract Group Theory 1.1 Group A group is a set of elements that have the following properties: 1. Closure: if a and b are members of the group, c = ab is also a member of the group. 2. Associativity: (ab)c = a(bc) for all a, b, c in the group. 3. Unit element: there is an element e such that ea = a for every element a in the group. 4. Inverse: to every element a there is a corresponding inverse element a 1 in the group such that a 1 a = e. The following properties follow from the above definition: 1. Left cancellation: If ax = ay then x = y for all a in the group. Proof: ax = ay a 1 (ax) = a 1 (ay) (a 1 a)x = (a 1 a)y ex = ey x = y. 2. Unit element on the right: ae = a = ea. Proof: a 1 (ae) = (a 1 a)e = ee = e = a 1 a and using the left cancellation law we have ae = a. 1

6 2 CHAPTER 1. ABSTRACT GROUP THEORY 3. Inverse element on the right: aa 1 = e = a 1 a. Proof: a 1 (aa 1 ) = (a 1 a)a 1 = ea 1 = a 1 = a 1 e. Using the left cancellation law, aa 1 = e. 4. Right cancellation: If xa = ya then x = y for all a in the group. Proof: xa = ya (xa)a 1 = (ya)a 1 x(aa 1 ) = y(aa 1 ) xe = ye x = y. We note the importance of associativity in the above proofs. The following identity is often useful: (ab) 1 = b 1 a 1 which follows from (ab) 1 (ab) = 1 (ab) 1 a = b 1 (ab) 1 = b 1 a Abelian Group If a group has a further property that ab = ba for all a, b in the group, the group is called Abelian. 1.3 Subgroup A subgroup is a set of elements within a group which forms a group by itself. Evidently, the unit element forms a subgroup by itself. We note that the definition of a group does not require all elements to be distinct from one another. However, when two elements are identical, then one of them is redundant so that we usually assume that all elements are different. A group with n identical elements evidently has at least n identical subgroups. 1.4 Examples 1. Integers under addition. The unit element e = 0 and the inverse of an element a is a 1 = a. This group is Abelian and infinite.

7 1.5. MULTIPLICATION TABLE 3 2. A set of all n n unitary matrices U (U = U 1 ) under matrix multiplication. The unit element is the unit matrix and the inverse of U is U by definition. We have to show that a product of two unitary matrices is unitary. (U 1 U 2 ) = U 2U 1 = U2 1 U1 1 = (U 1 U 2 ) 1 3. The set of permutation operations that take ABC into ABC, ACB, BAC, CAB, BCA, CBA. The elements of the group are e = ( A B ) C A B C α = ( A B ) C B C A β = ( A B ) C C A B γ = ( A B ) C A C B µ = ( A B ) C C B A ν = ( A B ) C B A C The operation αβ means: first do permutation β and then permutation α on the previous result. We show below that every permutation has an inverse permutation and two successive permutations correspond to a single permutation i.e. the permutations form a group. We see from the above examples that multiplication can mean addition, matrix multiplication etc. or simply that one operation is performed on the result of the preceeding operation like in the example Multiplication Table The group multiplication table is a matrix M ab = ab, where a and b are elements of the group and the matrix element corresponding to row a and column b is given by the group multiplication ab. We illustrate this definition by constructing the multiplication table of the permutation group in the example 3 above.

8 4 CHAPTER 1. ABSTRACT GROUP THEORY e α β γ µ ν e e α β γ µ ν α α β e µ ν γ β β e α ν γ µ γ γ ν µ e β α µ µ γ ν α e β ν ν µ γ β α e Thus, for example, the inverse of α is β and vice versa. The group is clearly not Abelian because e.g. αγ γα. The elements (e, γ) form a subgroup, so do (e, µ), (e, ν) and (e, α, β). Obviously, the unit element e forms a subgroup by itself. The group multiplication table mathematically characterises the group. All groups with the same multiplication table are mathematically identical with respect to their group properties. The multiplication table has the following properties: 1. Rearrangement Theorem: In every row or column, each element must appear once and once only and therefore each row (column) is different from any other row (column). Proof: Suppose element b appears twice in the row a. This means that ac = b and ad = b where c d. But this implies that ac = ad c = d which is a contradiction. Thus element b cannot appear more than once and since the size of the row or column is the same as the number of elements in the group, it follows that each element must appear once and once only. 2. The multiplication table is symmetric across the diagonal when the group is Abelian because the elements commute with one another. 1.6 Cyclic Groups A group of n elements is said to be cyclic if it can be generated from one element. The elements of the group must be a, a 2, a 3,..., a n = e

9 1.7. ORDER OF AN ELEMENT 5 n is called the order of the cyclic group. A cyclic group is evidently Abelian but an Abelian group is not necessarily cyclic. It is shown below that every non-cyclic group has at least a cyclic subgroup. Examples of cyclic groups are the subgroups of the permutation group in the example 3. The subgroup (e, α, β) is the same as (α, α 2 = β, α 3 = e). 1.7 Order of an Element Let a e be an element of a group. Form the products a 2, a 3,.... a 2 must be either e or a different element from a because if a 2 = a a = e. If a 2 e we continue forming a 3. By a similar argument, a 3 must be either e or a different element from a and a 2. If a, a 2,..., a n are distinct from each other and a n = e then n is called the order of element a. These elements form a cyclic group. Thus every group must have at least one cyclic subgroup. In the example 3 above, α and β are of order 3 and γ, µ, and ν are of order Properties of Finite Groups We summarise below the properties of finite groups. 1. Every element a has a finite order n such that a n = e. 2. Rearrangement Theorem: Multiplying all elements in a group by an arbitrary element reproduces the group. This has been proven above in the properties of the multiplication table. 3. Every non-cyclic group has at least a cyclic subgroup. 1.9 Cosets If S is a subgroup of G and a is an element of G not in S, then the sets as and Sa are called the left and right cosets of S respectively. a cannot be a unit element and therefore a coset can never be a group because it has no unit element. It is also evident that as or Sa have no element in common with S for otherwise, a should be included in S. Morevoer, if b is

10 6 CHAPTER 1. ABSTRACT GROUP THEORY an element of G which is neither in S nor in as, then bs has no element in common with either S or as. Proof: Let x and y be in S such that ax = by. We have axy 1 = b, but xy 1 is an element in S and therefore b is in as which is a contradiction. Consequently, a finite group can be factorised as G = S + as + bs +... but the factorisation is not unique. It depends on the choice of S, a, b,..., etc. The number of factors must be finite and the factorisation must exhaust the group. Thus we have the following theorem: Let h be the number of elements of a group G and g be the number of elements of a subgroup S of G. Then h g = integer As a consequence, if h is prime then G must be cyclic and it has no subgroup other than the trivial subgroup of e Class Two elements a and b of a group are conjugate to one another if there is an element g in the group such that a = gbg 1 (1.1) (Since every element has an inverse, it also follows that b = g ag 1 where g = g 1 ). The transformation gbg 1 is often called a similarity transformation. If a and b are conjugate to each other and b and c are conjugate to each other, then a and c are also conjugate to each other. This is because a = g 1 bg1 1 and b = g 2 cg2 1 a = g 1 g 2 cg2 1 g1 1 = g 1 g 2 c(g 1 g 2 ) 1. A class C is a set of elements which are conjugate to each other. The unit element evidently forms a class by itself. If G is Abelian, each element also forms a class by itself. It is clear that a group may be broken up into classes G = C 1 + C and an element cannot belong to more than one class.

11 1.11. ISOMORPHISM AND HOMOMORPHISM 7 As an example, we work out one of the classes of the permutation group. In working out the classes, it is obviously not necessary to consider similarity transformations with either the unit element or the element itself. We use the multiplication table that we have constructed previously: βαβ 1 = βαα = ββ = α γαγ 1 = γαγ = γµ = β µαµ 1 = µαµ = µν = β ναν 1 = ναν = νγ = β Thus α and β belong to the same class. Similarly we can show that (γ, µ, ν) also form a class. Elements belonging to the same class usually have the same characteristic. In the above example, α and β both correspond to cyclic permutations whereas γ, µ, and ν correspond to permutations with one member fixed Isomorphism and Homomorphism Two groups are said to be isomorphic if they have the same multiplication table, by reordering the elements if necessary. If F is isomorphic to G then there is a one-to-one correspondence between the elements of F and G, f i g i, such that if f i f j = f k then g i g j = g k. Two groups can only be isomorphic if they have the same number of elements. As an example, the permutation group in the example 3 is isomorphic to the symmetry operations that take an equilateral triangle into itself. This isomorphism can be seen by labelling the corners of the triangle with A, B, C. Homomorphism is similar to isomorphism except that the relationship is many-to-one. Two groups G and G are said to be homomorphic if each element of G can be associated with some elements of G : a a = (a 1, a 2,...), b b = (b 1, b 2,...), c c = (c 1, c 2,...), such that if ab = c then a ib j = c k i.e. for every i, j, c k lies in c.

12 8 CHAPTER 1. ABSTRACT GROUP THEORY 1.12 Invariant Subgroup We prove the following theorem: Theorem: If S is a subgroup of G and a is an element of G then S = asa 1 is also a subgroup and it is isomorphic to S. We assume that the elements of S are distinct. Proof: We first show that the elements of S are distinct. Let as i a 1 = as j a 1 and i j. Then by the cancellation law, s i = s j, which is a contradiction. Thus the elements of S are distinct. We now show that if s i s j = s k then s is j = s k. Proof: s is j = as i a 1 as j a 1 = as i s j a 1 = as k a 1 = s k. Thus S and S are isomorphic with the same unit element e = aea 1 = e and the inverse of s i is s 1 i = (as i a 1 ) 1 = as 1 i a 1 = (s 1 i ). If S is the same as S for every a in the group then the subgroup S is called an invariant subgroup. This implies that the left coset of S is the same as its right coset, i.e. as = Sa If we break a group into its cosets, G = S + as + bs,... with an invariant subgroup S, then these cosets form a group with the unit element equal to S. Thus asbs = abss = abs = cs if ab = c. This group is called a factor group. It is an example of homomorphism where the invariant subgroup S is associated with the elements of S and the coset as is associated with the elements of as Direct Product Group Let F be a group with elements f i, i = 1,..., h F, and G be a group with elements g i, i = 1,..., h G, such that f i g j = g j f i for all i and j. The direct product group F G is defined to be the set of all distinct elements f i g j. If F and G have no common element, apart from the identity, then the order of the direct product group will be h F h G. We show that F G is a group.

13 1.13. DIRECT PRODUCT GROUP 9 1. Closure: (f i g j )(f k g l ) = (f i f k )(g j g l ) = f r g s. Since f r is in F and g s is in G then by definition f r g s is in F G. 2. Unit element is e F e G. 3. Inverse of f i g j is (f i g j ) 1 = gj 1 fi 1 4. Associativity: obvious. We have the following theorem: = fi 1 gj 1. Theorem: The classes of the direct product group are given by the direct products of the classes of the individual groups. Proof: We label the elements of the product group a ij = f i g j. According to the definition of a class, we have C ij = a 1 rs a ij a rs, for all r, s = fr 1 gs 1 f i g j f r g s = (fr 1 f i f r )(gs 1 g j g s )

14 10 CHAPTER 1. ABSTRACT GROUP THEORY

15 Chapter 2 Theory of Group Representations 2.1 Definitions A group {A, B, C,...} may be represented by a set of square matrices {T (A), T (B), T (C),...}. These matrices are said to form a representation for the group if they satisfy the same group multiplication rule: AB = C T (A)T (B) = T (C) (2.1) The identity element E is represented by a unit matrix and each matrix must have an inverse. Some or all of the matrices may be the same. If the matrices are different, the representation is called faithful (isomorphic). The order of the matrices is called the dimension of the representation. 2.2 Equivalent Representations If T is a representation then T = S 1 T S (2.2) with an arbitrary but non-singular S is also a representation because T (A)T (B) = S 1 T (A)SS 1 T (B)S 11

16 12 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS = S 1 T (A)T (B)S = S 1 T (AB)S = T (AB) T and T are equivalent and there are an infinite number of equivalent representations. 2.3 Reducible and Irreducible Representation A representation T is reducible if there exists a non-singular matrix S such that the equivalent representation T = S 1 T S has the form T 1 T 2 T = T 3. for all elements in the group. T 1, T 2, T 3,... are square matrices and the rest of the elements are zero. The representation is irreducible if it cannot be reduced into the above form Equivalent Unitary Representation There are an infinite number of equivalent representations but it is possible to find one which is unitary. Theorem 1: Given an arbitrary representation T, there is always an equivalent unitary representation. Proof: Construct a Hermitian matrix H = T (G)T (G) G

17 2.4. EQUIVALENT UNITARY REPRESENTATION 13 According to matrix algebra, we can always diagonalise a Hermitian matrix by a unitary transformation D = U 1 HU = G U 1 T (G)T (G)U = G = G U 1 T (G)UU 1 T (G)U T (G)T (G) U is made up of the eigenvectors of H and D is diagonal by definition and its diagonal elements are real and positive. D jj = G = G = G k k k T jk(g)t kj(g) T jk(g)t jk(g) T jk(g) 2 0 In fact, it can never be equal to zero because otherwise T = 0. Since D is diagonal, we may define a diagonal matrix D 1/2 with diagonal elements equal to D 1/2 ii so that 1 = D 1/2 G T (G)T (G)D 1/2 Using the above result, the matrix T (G) = D 1/2 T (G)D 1/2 can be shown to be unitary. T (G)T (G) = D 1/2 T (G)D 1/2 [D 1/2 G T (G )T (G )D 1/2 ] D 1/2 T (G)D 1/2 = D 1/2 G T (G)T (G )T (G )T (G)D 1/2 = D 1/2 G T (G)T (G )[T (G)T (G )] D 1/2 = D 1/2 G T (GG )T (GG )D 1/2

18 14 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS = D 1/2 G T (G )T (G )D 1/2 = 1 The second last step has been obtained by using the Rearrangement Theorem. Thus, given an arbitrary representation T, it is always possible to construct a unitary representation by forming T = D 1/2 U 1 T UD 1/2 (2.3) From now on we assume that we have a unitary representation. 2.5 Schur s Lemma Schur s lemma is used in proving many of the theorems in group theory. Theorem 2 (Schur s lemma): Given that [M, T ] = MT T M = 0 for all elements in the group, then (a) if T is irreducible, M = ci where c is a constant and I is a unit matrix. (b) if M ci then T is reducible. Proof: We first show that it is sufficient to prove the theorem for a Hermitian M. T M = MT (T M) = (MT ) M T = T M T (M T )T = T (T M )T T M = M T since T = T 1 (unitary). Adding and substracting the first and the last equations, we get T (M + M ) = (M + M )T and T i(m M ) = i(m M ) T Let H 1 = M + M and H 2 = i(m M ) so that [H 1, T ] = [H 2, T ] = 0. H 1 and H 2 are Hermitian and M = 1/2(H 1 ih 2 ). Thus, if the theorem is

19 2.5. SCHUR S LEMMA 15 true for a Hermitian matrix, it must be true for M. We can now assume M to be Hermitian and perform a unitary transformation so that M becomes diagonal, D = U 1 MU, and define an equivalent representation T = U 1 T U. Then T D = U 1 T UU 1 MU = U 1 T MU = U 1 MT U = U 1 MUU 1 T U = DT We have to show that the diagonal elements of D are all identical if T, or equivalently T, is irreducible. Consider the ij, i j, element, T ijd jj = D ii T ij T ij(d jj D ii ) = 0 (2.4) Let us order the diagonal elements of D such that D ii D jj for i < j which can always be done by rearranging the columns of U. Suppose D 11 = D nn and D 11 < D ii where n 1 and i > n. Then it follows from Eq. ( 2.4) that T must have the form T = ( X 0 0 Y where X is an n n matrix, i.e. T is reducible. Let us summarise the conclusions: If M ci then T must be reducible. If M = ci then obviously it commutes with T whether it is reducible or irreducible. Hence if T is irreducible then M must be equal to ci for otherwise T would be reducible. Theorem 3: Given T α M = MT β, where T α and T β are irreducible representations of dimension l α and l β respectively and M is a rectangular l α l β matrix, then (a) if l α l β, M = 0 or (b) if l α = l β either M = 0 or M = 0. In the latter case, M will have an )

20 16 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS inverse so that T α and T β are equivalent. Proof: T α (G)M = MT β (G) (T α (G)M) = (MT β (G)) M T α (G) = T β (G)M M T α 1 (G) = T β 1 (G)M MM T α 1 (G) = MT β 1 (G)M MM T α (G 1 ) = MT β (G 1 )M MM T α (G 1 ) = T α (G 1 )MM Since T α is irreducible, it follows from Schur s lemma (Theorem 2) that MM = ci. Consider the case l α = l β. MM = ci ; M M = M 2 = c lα If c 0, then M = 0 i.e. M 1 exists. It follows that T α and T β are equivalent since M 1 (T α (G)M) = M 1 (MT β (G)) = T β (G) If c = 0, then MM = 0. Let us look at the diagonal elements of MM : M ik M ki = 0 k k k M ik M ik = 0 M ik 2 = 0 Since this is positive definite, each term must vanish i.e. M = 0. Consider now the case l α l β, l α > l β. We enlarge M into a l α l α square matrix N with the additional elements equal to zero. Then NN = MM = ci. But N = 0 because one of its columns is zero. NN = ci NN = c lα N N = c lα c = 0 Hence NN = 0 which implies N = 0. Since M is contained in N, then M = 0.

21 2.6. ORTHOGONALITY THEOREM Orthogonality Theorem We are now in a position to prove the Orthogonality Theorem which is central in representation theory. Theorem 4: Orthogonality Theorem G T α ij(g) T β kl (G) = (h/l α)δ αβ δ ik δ jl (2.5) α and β label the irreducible representations, l α and l β are the dimensions of these irreducible representations, and h is the number of elements in the group. Proof: Define a matrix M = G T α (G)XT β (G 1 ) where X is an arbitrary matrix and α β. We want to show that this matrix satisfies the postulates of Theorem 3. Consider T α (A)M = G T α (A)T α (G)XT β (G 1 ) = G = G = G = G T α (A)T α (G)XT β (G 1 )T β (A 1 )T β (A) T α (AG)XT β (G 1 A 1 )T β (A) T α (AG)XT β ((AG) 1 )T β (A) T α (AG)XT β 1 (AG)T β (A) = [T α (G)XT β 1 (G)]T β (A) G = MT β (A) We have again made use of the Rearrangement Theorem in the second last step. According to Theorem 3, M = 0 since we are considering the case of α β so that M ij = 0 = G kl T α ik(g)x kl T β lj (G 1 )

22 18 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS Since X is arbitrary, we may set X kl = δ km δ ln. Then Tim(G)T α nj(g β 1 ) = Tim(G)T α β 1 nj (G) G G = G T α im(g)t β nj(g) = G = 0 T α im(g)t β jn(g) When α = β, according to Schur s lemma (Theorem 2) M ij = c δ ij = Tik(G)X α kl Tlj α (G 1 ) G = G kl T α im(g)t α nj(g 1 ) Putting i = j and summing over i yields c l α = Tim(G)T α ni(g α 1 ) G i = Tnm(E) α G Thus we have Since T is unitary, G G = hδ nm T α im(g)t α nj(g 1 ) = (h/l α )δ nm δ ij T α im(g)t α jn(g) = (h/l α )δ nm δ ij Eqn. ( 2.5) has the form of a dot product with Tij(G) α as the G th component of a vector labelled by α, i and j, in the h dimensional space of the group elements. This means that the number of distinct labels (α, i, j) cannot exceed h, i.e. α lα 2 h. It will be proven later that in fact lα 2 = h α The result is very useful in working out the irreducible representations of a group.

23 2.7. THE CHARACTERS OF A REPRESENTATION The Characters of a Representation We have seen that a representation is not unique. We recall that if T is a representation, so is T = S 1 T S. Given T and T, how can we then tell that they are equivalent? One way will be to figure out if there is a matrix S, such that T = S 1 T S. But this is a complicated procedure. What we are looking for are properties of a matrix which are invariant under a similarity transformation. The eigenvalues are one of them but they are too cumbersome to calculate. A simpler quantity is the trace of a matrix or the sum of the diagonal elements: χ α (G) = T r T α (G) = i T α ii (G) (2.6) We show that the trace or character of a matrix is invariant under a similarity transformation: T r T = Sij 1 T jk S ki i jk = ( jk i S ki S 1 ij )T jk = jk δ kj T jk = T r T The characters of a representation α is a set of h numbers {χ α (G)}. It will be shown later that if the representation is irreducible, its characters are unique. Theorem 5: If elements A and B of a group belong to the same class, then the characters of their representations are the same. Proof: Since A and B belong to the same class, there is an element C such that A = C 1 BC. Consequently, T r T (A) = T r [T (C 1 ) T (B) T (C)]

24 20 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS = T r [T (C) T (C 1 ) T (B)] = T r [T (E) T (B)] = T r T (B) Theorem 6: First Orthogonality Relation χ α (G) χ β (G) = χ α (C k ) χ β (C k )N k = hδ αβ G k (2.7) C k labels a class with N k elements. Proof: Setting i = j and k = l in Eqn. ( 2.5) (Orthogonality Theorem) and summing over i and l we have χ α (G) χ β (G) = (h/l α )δ αβ δ il δ il G il χ α (C k ) χ β (C k )N k = hδ αβ k Eqn. ( 2.7) has the form of a weighted dot product with weight N k in a space with dimension equal to the number of classes of the group. χ α (C k ) is the k th component of vector α. It follows that the number of irreducible representations must be less than or equal to the number of classes. 2.8 The Regular Representation We prove the following theorem: Theorem 7: α l 2 α = h (2.8) Proof:

25 2.8. THE REGULAR REPRESENTATION 21 A proof of this theorem is provided by considering the so called regular representation. The regular representation is formed in the following way. We write the multiplication table as follows: e α 1 β 1 γ 1 e α β γ... e e e e We form the regular representation of an element a by writing 1 wherever the element a occurs in the table and zero otherwise. Let the group elements be a i, i = 1, 2,..., h, then the regular representation is given by T jk (a i ) = 1 if a 1 j a k = a i = 0 otherwise It is clear from the definition that the representation is faithful. To show that it is really a representation, we have to prove that T jk (a i )T kl (a m ) = T jl (a p ) k if and only if a i a m = a p. By definition, T jk (a i ) 0 only when k is such that a k = a j a i and similarly T kl (a m ) 0 only when k is such that a k = a l a 1 m. Therefore k T jk (a i )T kl (a m ) = 1 if and only if a k = a j a i = a l a 1 m and zero otherwise. However, a j a i = a l a 1 m implies a 1 j a l = a i a m = a p. The character is given by h χ(a i ) = T jj (a i ) = h if a i = e j=1 = 0 otherwise This is clear by inspection of the multiplication table. We can now prove Theorem 7 by writing the character of the regular representation as a sum over characters of the irreducible representations: χ(a i ) = m α χ α (a i ) α

26 22 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS From the orthogonality relation we have m α = 1 χ α (a i )χ(a i ) h i Since χ α (e) = l α, χ(e) = h, and χ(a i e) = 0 it follows that m α = l α. Each irreducible representation occurs in the regular representation a number of times equal to the dimension of the irreducible representation. On the other hand, h = α m α l α so that h = α l 2 α This result together with the orthogonality theorem show that there are exactly h orthogonal vectors T α ij. Theorem 8: Second Orthogonality Relation χ α (C k ) χ α (C l ) = (h/n k )δ kl (2.9) α Proof: We recall that the matrix representation Tij α can be regarded as a set of orthogonal vectors in the h- dimensional space of the group elements. Tij(G) α is the component of the vector Tij α along the G-axis. Tij α themselves can be regarded as a basis in the space of the group elements since the number of α, i, j is precisely h. Any vector in this space can therefore be expanded in T α ij: χ = αij c αij T α ij We sum the components of χ along G-axes which belong to a given class C k, χ(c k ) = χ(g) G C k = G C k αij = 1 h G 1 G C k αij c αij T α ij(g) c αij T α ij(g 1 1 GG 1 )

27 2.8. THE REGULAR REPRESENTATION 23 = 1 c αij T h ik(g α 1 1 )Tkl(G)T α lj α (G 1 ) G 1 G C k αij kl = c αij δ ij δ kl Tkl(G)/l α α G C k αij kl = = α G C k αi a α χ α (C k ) c αii χ α (G)/l α where a α = (N k /l α ) i c αii. The third step uses the fact that G 1 1 C k G 1 = C k for every G 1 in the group, the fifth step uses the orthogonality theorem, and the last step uses the fact that the characters of elements belonging to the same class are identical. Thus the set of vectors {χ(c k )} are linearly independent because the group elements can be divided uniquely and completely into distinct classes. Multiplying χ(c k ) by χ β (C k )N k and summing over the classes k yields χ(c k )χ β (C k )N k = a α χ α (C k )χ β (C k )N k k α k = α a α hδ αβ = ha β Consequently, χ(c l ) = β = (1/h) β a β χ β (C l ) χ(c k )χ β (C k )N k χ β (C l ) k 0 = χ(c k ) (N k /h) χ β (C k )χ β (C l ) δ kl k β This is true for an arbitrary χ(c k ) so that the quantity in the square bracket must vanish which proves the theorem. Eqn. ( 2.9) has the form of a dot product in a space with dimension equal to the number of irreducible representations. It follows that the number of classes must be less than or equal to the number of irreducible representations. On the other hand by Theorem 6 the number of irreducible

28 24 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS representations must be less than or equal to the number of classes. Thus we conclude the following corollary: The number of irreducible representations is equal to the number of classes. Theorem 6 and 8 imply that the irreducible representations are uniquely characterised by their characters and consequently two distinct irreducible representations cannot have the same set of characters. 2.9 The Character Table The character table is a square matrix with dimension equal to the number of classes or irreducible representations: It has the form C 1 = C E N 2 C 2 N 3 C 3... T T 2 l T 3 l Q αk = χ α (C k ) The row is labelled by the irreducible representations and the column by the classes Properties of the Character Table The properties listed below are useful for constructing the character table. 1. The number of irreducible representations is equal to the number of classes. 2. The first column contains the dimension of each representation and the sum of its square is the total number of elements in the group (Theorem 7): α l 2 α = h.

29 2.10. PROPERTIES OF THE CHARACTER TABLE As a convention, the first row is the identity or unit representation consisting of Each row or column can be regarded as a vector. In this sense: (a) The rows are orthogonal with weighting factor N k (Theorem 6) and normalised to h. (b) The columns are orthogonal and normalised to h/n k (Theorem 8). 5. If the character table of a factor group is known, then due to homomorphism, the character of an element as in the factor group may be assigned to the classes which are contained in as. We recall that if a group has an invariant subgroup S, then we can form a group S, as, bs,... where a is not in S, b is not in S and as etc. This group is a called a factor group with S as the unit element and it is homomorphic with the group itself. An invariant subgroup means that x 1 Sx = S for every element x in the group and by Theorem 9 discussed in the next section, S must consist wholly of classes. The homomorphism between the factor group and the group itself means that if T (as) is a matrix representation of the element as of the factor group then the same matrix may be used to represent all elements in the coset as. Thus, the characters of the classes in as is just the same as the character of T (as). The above five rules are often sufficient to construct the character table but the following rule, which is described in the next section, can be of help. 6. The characters of the α representation are related by N i χ α (C i )N j χ α (C j ) = l α m ijk N k χ α (C k ) (2.10) where m ijk is the coefficients of class multiplication, C i C j = k m ijk C k. k

30 26 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS 2.11 Class Multiplication Let C be a collection of classes. From the definition of a class we have x 1 Cx = C for every element x in the group. This is evident from the fact that all elements in a class are conjugate to one another and if two elements a and b are different then x 1 ax x 1 bx. Moreover, an element cannot belong to more than one class. Thus the elements on the right hand side must be identical to the elements in C. Theorem 9: A set of elements C obeying x 1 Cx = C for all x in the group must be composed wholly of classes. Proof: Suppose R is a set of elements in C which do not form a class. But x 1 Rx must be equal to R itself since (C R) form classes and by definition they will never be conjugate to elements in R. Thus R must be a class. We now consider a product of two classes and here we do count all resulting elements even when some are the same. C i C j = x 1 C i xx 1 C j x = x 1 C i C j x By Theorem 9, it follows that C i C j must consist wholly of classes. Therefore it must be possible to write C i C j = k m ijk C k (2.11) where m ijk are integers telling how often the class C k appears in the product C i C j. Let us consider the class multiplication in terms of matrix representations. If we let S(C i ) be the sum of matrices of all elements in the class C i and T (x) be a matrix representation of the element x then we have T 1 (x)s(c i )T (x) = S(C i ) or S(C i )T (x) = T (x)s(c i ). If the representation T is irreducible, then it follows from Schur s Lemma that S(C i ) = c i I.

31 2.12. DIRECT PRODUCT GROUPS 27 Thus c i c j = k m ijk c k Taking the trace of S(C i ) and assuming we are in the irreducible representation α we get T r S(C i ) = T r c i I = c i l α On the other hand, Therefore and Eq. ( 2.10) follows. T r S(C i ) = N i χ α (C i ) c i = N i χ α (C i )/l α 2.12 Direct Product Groups If a group is a direct product of two groups, then the irreducible representations and the character table of the product group can be worked out easily from those of the two individual groups with the help of the following theorem. Theorem 10: If G = G 1 G 2 then T α β = T1 α T β 2 is an irreducible representation of G. Moreover, for all T1 α and T β 2 these are all possible irreducible representations of the product group. Proof: We first show that T is a representation of G. Let a 1, a 2,... and b 1, b 2,... be the elements of G 1 and G 2 respectively and let c ij = a i b j be the elements of the direct product group G. Suppose c ij c kl = c mn which implies that a i b j a k b l = a m b n or a i a k b j b l = a m b n since the elements of G 1 and G 2 commute. This means that a i a k = a m and b j b l = b n or T α (a i )T α (a k ) = T α (a m ) and T β (b j )T β (b l ) = T β (b n ). Writing the last two matrix equations in component form, multiplying them and using the definition of direct product yield the required result. Let M be a matrix which commutes with every T α β. M may be written in the form M = M 1 M 2 where M 1 and M 2 have the same dimensions as T1 α and T β 2 respectively. The commutivity of M with T α β

32 28 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS implies the commutivity of M 1 with T1 α and M 2 with T β 2. By Schur s lemma, M 1 = k 1 I 1 and M 2 = k 2 I 2 and hence M = k 1 k 2 I and no nonconstant matrix M exists so that by Schur s lemma T α β is irreducible. To show the last part of the theorem, we recall that α l1α 2 = h 1 and β l2β 2 = h 2. Let l αβ = l1 α l β 2 be the dimension of the irreducible representation T α β. We have lαβ 2 = l 2 1α l2β 2 = h 1 h 2 α αβ But h 1 h 2 is the order of the direct product group and therefore there are no more possible irreducible representations other than T α β for all T1 α and T β Examples 1. We construct the character table of the permutation group of three objects which we have considered previously. There are three classes C 1 = e, C 2 = (α, β), and C 3 = (γ, µ, ν) and therefore by rule 1 there are three irreducible representations. Rule 2 tells us that l l l 2 3 = 6. There is always an identity representation so we may take l 1 = 1. Then l 2 = 1 and l 3 = 2 are the only possible solution and the character table must look like as follows: β C 1 2C 2 3C 3 T T 2 1 a b T 3 2 c d To determine a, b, c, and d we need to know one of them and we can find the others by orthogonality (rule 4). The subgroup S = (e, α, β) = C 1 + C 2 is an invariant subgroup and the corresponding factor group has the multiplication table S γs S S γs γs γs S

33 2.13. EXAMPLES 29 where γs = µs = νs = (γ, µ, ν) = C 3. This is a group of two elements and there are evidently two classes, the first class consists of the unit element S and the second class consists of the element γs. The number of irreducible representations is two and both must have dimension one, by rule 2. The irreducible representations of the factor group must be (1, 1) and (1, 1). Then from rule 5 we may assign a = 1 and b = 1 (a = 1 and b = 1 gives the identity representation). Note that we have taken a and b and not c and d because the irreducible representations of the factor group are one dimensional and T 2 is also a one dimensional representation. The other two characters c and d can be easily obtained from the orthogonality between columns. Finally we have a + 2 c = 0 c = b + 2 d = 0 d = 0 C 1 2C 2 3C 3 T T T Abelian groups: Every element in an Abelian group forms a class by itself. Therefore the number of irreducible representations is simply equal to the number of elements in the group. Moreover, rule 2 implies that all irreducible representations have dimension one. The consequences are (a) Unitarity of representations implies that the numbers representing an Abelian group have modulus one.

34 30 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS (b) If the number X represents an element x then X represents x 1. (c) The character table serves also as a representation table. As an example we consider a non-cyclic group of fourth order with the following multiplication table e a b c e e a b c a a e c b b b c e a c c b a e a 2 = b 2 = c 2 = e so a, b, and c are represented either by ±1. If a is represented by 1 then b = c = ±1 because bc = a. If a = 1 then either b = 1 and c = 1 or b = 1 and c = 1. Thus we have e a b c T T T T Cyclic groups: These are Abelian groups which can be generated by a single element: G = (a, a 2,..., a h = e). If a is represented by a number A k then a 2 is represented by A 2 k etc. and A h k = 1 since e is always represented by 1. Thus, A k must be one of the h roots of

35 2.13. EXAMPLES 31 unity, i.e. A k = exp(i2πk/h) and the character table is given by e a a 2... a h 1 T T 2 1 A 1 A A h 1 1 T 3 1 A 2 A A2 h T h 1 A h 1 A 2 h 1... Ah 1 h 1

36 32 CHAPTER 2. THEORY OF GROUP REPRESENTATIONS

37 Chapter 3 Group Theory in Quantum Mechanics 3.1 Linear Vector Space A set {r 1, r 2,...} is said to form a linear vector space L if r i + r j is another member in the set for every i, j and cr i is also another member in the set where c is a complex constant. The quantities {r 1, r 2,...} are called vectors and there are an infinite number of them. A set of vectors {v 1, v 2,..., v p } is said to be linearly independent if none of them can be expressed as a linear combination of the others. If a set of coefficients {c k } can be found such that p c k v k = 0 k=1 then the vectors are said to be linearly dependent. The largest number of vectors in L which form a linearly independent set is called the dimension of L. This is the same as the smallest number of vectors needed to describe every vector in L and these vectors are said to form a basis. We denote the dimension of L by s and the basis vectors by {e 1, e 2,..., e s } so that any vector v may be written as s v = v i e i i=1 33

38 34 CHAPTER 3. GROUP THEORY IN QUANTUM MECHANICS In order to determine the coefficients v i from a given vector v we introduce the concept of scalar or dot product between two vectors v 1 and v 2 which we denote by (v 1, v 2 ). The particular definition of the scalar product is arbitrary but it must satisfy the following general conditions: 1. (v 1, v 2 ) is a complex number 2. (v 1, v 2 ) = (v 2, v 1 ) 3. (v 1, cv 2 ) = c(v 1, v 2 ) (cv 1, v 2 ) = (v 2, cv 1 ) = c (v 1, v 2 ) 4. (v 1 + v 2, v 3 ) = (v 1, v 3 ) + (v 2, v 3 ) 5. (v 1, v 1 ) 0. It is only zero when v 1 = 0. Examples: 1. Vectors in 3-dimensional space. The dot product is defined as (v 1, v 2 ) = v 1 v 2 cosθ where v 1 and v 2 are the lengths of the vectors and θ is the angle between the two vectors. 2. Function spaces. The dot product is defined as (ψ i, ψ j ) = d 3 r w(r)ψ i (r)ψ j (r) where w is a weight function which is usually real and positive definite so that the last condition is satisfied. We can also define orthonormal basis functions {φ i } such that any function can be expanded in this basis: ψ = i=1 c i φ i The linear space spanned by this basis is called the Hilbert space. We define an operator ˆT by ˆT v = v where v is an arbitrary vector or function which is carried into another vector or function v. Both v and v are in L. The operator is called linear if ˆT (v 1 + v 2 ) = ˆT v 1 + ˆT v 2 ˆT cv = c ˆT v

39 3.2. SYMMETRY TRANSFORMATIONS 35 We will consider only linear operators. Since any vector in L can be expanded in the basis vectors, it is only necessary to study the effects of the operator on the basis vectors. ˆT e j = i e i T ij The matrix T ij is said to form a representation for the operator ˆT in the linear vector space L. It is usually convenient to choose an orthonormal basis, by this we mean (φ i, φ j ) = δ ij. This can always be done by the Gramm-Schmidt orthogonalisation procedure. We illustrate the method for three vectors. The generalisation to an arbitrary number of vectors is straightforward. Let φ 1, φ 2, φ 3 be three linearly independent functions which are not necessarily orthogonal nor normalised. Let ψ 1 = φ 1 / (φ 1, φ 1 ) and construct ψ 2 = φ 2 ψ 1 (ψ 1, φ 2 ) which is orthogonal to ψ 1. We normalise ψ 2 to get ψ 2 = ψ 2 / ( ψ 2, ψ 2 ) and construct ψ 3 = φ 3 ψ 2 (ψ 2, φ 3 ) ψ 1 (ψ 1, φ 3 ) which is orthogonal to both ψ 1 and ψ 2 and which is then normalised to give ψ 3 = ψ 3 / ψ3, ψ 3 ) In this way we have constructed an orthonormal set of functions and we can see that for a general case we have ψ n = φ n n 1 i=1 ψ i (ψ i, φ n ) 3.2 Symmetry Transformations A symmetry transformation G with respect to a given function f(x, y, z) is a real linear coordinate transformation that preserves the length (real unitary transformation) r = Gr r i = j G ij r j, r i = x, y, z such that the function calculated with the new variables r i by the above substitution of coordinates is the same as the function calculated with the old variables: f(x, y, z ) = f (x, y, z) = f(x, y, z) f(gr) = f(r). In

40 36 CHAPTER 3. GROUP THEORY IN QUANTUM MECHANICS other words, the function looks the same in the old and the new coordinate system. This also implies that f(gr) = f(r) = f(g 1 r) because f(r ) = f(r) = f(g 1 r ) and r is just a dummy variable. A symmetry operator ˆT (G) associated with a symmetry transformation G is defined by ˆT (G)f(r) = f(g 1 r) It is important to note that the operator acts upon the coordinates r, and not upon the argument of f. Thus we mean that ˆT (G)f(Ar) = f(ag 1 r) f(g 1 Ar) We also have G 1 instead of G in our definition. This is a natural choice because two successive operations ˆT (G 1 ) and ˆT (G 2 ) on a function f(r) correspond to the same order of coordinate transformation: ˆT (G 2 ) ˆT (G 1 )f(r) = ˆT (G 2 )f(g 1 1 r) = f(g 1 1 G 1 2 r) = f([g 2 G 1 ] 1 r) Theorem 1: If {G} is a set of all symmetry transformations of a function, then the associated symmetry operators { ˆT (G)} form a group. Proof: The set {G} evidently form a group because if G is a symmetry transformation, then so is G 1 as discussed above. A succession of two symmetry transformations is itself a symmetry transformation and the identity transformation is obviously a symmetry transformation. From the definition, we have ˆT (G 1 ) ˆT (G 2 )f(r) = ˆT (G 1 )f(g 1 2 r) = f(g 1 2 G 1 1 r) = f([g 1 G 2 ] 1 r) = ˆT (G 1 G 2 )f(r) Hence ˆT (G 1 ) ˆT (G 2 ) = ˆT (G 1 G 2 ) and the symmetry operators form a group.

41 3.3. INVARIANT SUBSPACE AND ITS GENERATION Invariant Subspace and Its Generation A set of functions or vectors {φ i } is said to span an invariant subspace V S under a given set of symmetry operators { ˆT (G)} if ˆT (G)φ i is also in V S for all G and i. An invariant subspace can be generated from an arbitrary function f and a set of symmetry operators { ˆT (G)} by applying each of the operators on f. The set of functions { ˆT (G)f} form an invariant subspace under { ˆT (G)} because ˆT (G )[ ˆT (G)f] = ˆT (G G)f, which must lie in the subspace since ˆT (G G) is just another member of { ˆT (G)}. We note that the number of functions that span the subspace is not necessarily equal to the number of G. Some of the functions may be the same or can be expressed as linear combinations of the others. The largest number of linearly independent functions in V S is called the dimension of the subspace. It may happen that we can construct a function f in V S such that the invariant subspace { ˆT (G)f } has a smaller dimension than that of V S. In this case the subspace V S is said to be reducible. When such a function cannot be found than V S is said to be irreducible. 3.4 Basis Functions for a Representation Let {φ i } be a set of orthonormal functions that span an invariant subspace V S. Then ˆT (G)φ i (r) = φ i (G 1 r) is in V S by definition and it can therefore be expanded as a linear combination of {φ i }: ˆT (G)φ i = j φ j T ji (G) (3.1) If the basis functions are orthonormal, then T (G) is unitary because ( ˆT (G)φ i, ˆT (G)φ j ) = δ ij = ( k φ k T ki (G), l φ l T lj (G)) = kl = k T ki(g)t lj (G)(φ k, φ l ) T ik (G)T kj(g)

42 38 CHAPTER 3. GROUP THEORY IN QUANTUM MECHANICS Theorem 2: The matrices {T (G)} form a representation for { ˆT (G)} Proof: ˆT (G 1 ) ˆT (G 2 )φ i = j ˆT (G 1 G 2 )φ i = jk φ k T ki (G 1 G 2 ) = k k ˆT (G 1 )φ j T ji (G 2 ) φ k T kj (G 1 )T ji (G 2 ) φ k T kj (G 1 )T ji (G 2 ) Thus T (G 1 )T (G 2 ) = T (G 1 G 2 ). The functions {φ i } are said to form a basis for a representation, which may be reducible. Evidently, if the subspace is irreducible under { ˆT (G)}, then the representation is also irreducible. This simply follows from the fact that one always generates the entire subspace starting from any function in the subspace. If the representation were reducible, then it would be possible to start from a function that does not generate the entire subspace, in which case the subspace is reducible. We now see the relevance of matrix representation theory studied in the previous chapter. Since the theorems were developed for arbitrary matrices, they are also applicable here. Theorem 3: A change of basis vectors ψ j = i φ i S ij corresponds to a similarity transformation of the original representation, which is an equivalent representation. S is assumed to be unitary so that (ψ i, ψ j ) = δ ij. Proof: ˆT (G) j ˆT (G)φ i = j ψ j S 1 ji = j φ j T ji (G) k ψ k S 1 kj T ji(g)

43 3.5. PROJECTION OPERATORS 39 Multiplying both sides by S im and summing over i yields ˆT (G)ψ m = k ψ k T km(g) where T km(g) = ji S 1 kj T ji(g)s im (G) or T = S 1 T S which is a similarity transformation. It is clear that the reverse is also true, i.e. a similarity transformation of the representation with a unitary matrix S corresponds to a unitary transformation of the basis vectors. 3.5 Projection Operators It is possible to construct a projection operator ˆP β such that given an arbitrary function ψ, ˆP β ψ is a component of ψ in the irreducible subspace β, i.e. ˆP β ψ transforms according to the irreducible representation β. This is a very useful tool. In the following, we construct these projection operators. Suppose {ψ α i } span an irreducible subspace transforming according to the irreducible representation α of { ˆT (G)}. That is ˆT (G)ψ α i = j ψ α j T α ji(g) Multiplying by T β kl G (G) and summing over G yields T β kl (G) ˆT (G)ψ α i = G j ψ α j T α ji(g)t β kl (G) Using the Orthogonality Theorem on the right side, we get G T β kl (G) ˆT (G)ψ α i = j ψ α j (h/l β )δ jk δ il δ αβ = (h/l β )δ il δ αβ ψ α k We define ˆP β kl (l β/h) G T β kl(g) ˆT (G) so that ˆP β kl ψα i = δ il δ αβ ψ β k ˆP α kiψ α i = ψ α k (3.2)

44 40 CHAPTER 3. GROUP THEORY IN QUANTUM MECHANICS Starting from a given function that belongs to the ith row of a given irreducible representation α, the projection operator can be used to generate the other partner functions corresponding to the other rows. These functions form a basis for the irreducible representation α. Moreover, we can also define ˆP β i ˆP β ii (l β /h) G χ β (G) ˆT (G) so that ˆP β ψ α i = δ βα ψ α i (3.3) The projection operators allow us to decompose the space, in which the symmetry operators are defined, into irreducible subspaces. Any function ψ can therefore be written as follows ψ = α ψi α (3.4) i where ψi α belongs to the ith row of the irreducible representation α. We note also that if φ α i and ψi α belong to the ith row of the irreducible representation α, so does any linear combination aφ α i + bψi α. Suppose that a given representation T is reducible. Then there is a unitary matrix S such that T = S 1 T S = α m α T α has the block form and each block cannot be reduced any further. α labels the irreducible representation and m α tells us how many times the irreducible representation α appears in the reduction. We note that when m α > 1 the irreducible representations T α are assumed to be the same. This can always be done by means of similarity transformations in each subspace which transforms according to the same irreducible representation α. The new basis vectors according to Theorem 3 are given by ψ j = i φ i S ij and they may be labelled according to the irreducible subspaces to which they belong. Any vector in V S can therefore be expanded as follows: v = α,t,i ψi αt c αt i = vi α (3.5) α,i